The equation of the normal to the curve `y=x(2-x)` at the point `(2,\ 0)` is `x-2y=2` (b) `x-2y+2=0` (c) `2x+y=4` (d) `2x+y-4=0`

Explanation

The given curve is
` y=2x−x^2
`

i.e, `x^2−2x+y=0`


Now the equation of the tangent at the point `(x_1​,y_1​)=(2,0)`

or, `x x_1​−(x+x_1​)+1/2​(y+y_1​)=0`

i.e `2x+(−1)(x+2)+1/2​(y+0)=0`

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